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The first couple of minutes of the launch of a space shuttle can be described very roughly as follows: The initial mass is \(2 \times 10^{6} \mathrm{kg}\), the final mass (after 2 minutes) is about \(1 \times 10^{6} \mathrm{kg}\), the average exhaust speed \(v_{\mathrm{ex}}\) is about \(3000 \mathrm{m} / \mathrm{s},\) and the initial velocity is, of course, zero. If all this were taking place in outer space, with negligible gravity, what would be the shuttle's speed at the end of this stage? What is the thrust during the same period and how does it compare with the initial total weight of the shuttle (on earth)?

Short Answer

Expert verified
Speed is 2079 m/s, thrust is 25,000,000 N, and it's 1.27 times the initial weight.

Step by step solution

01

Understand the Rocket Equation

The rocket equation, or Tsiolkovsky's rocket equation, is used to estimate the final velocity (\(v_f\)) of a spacecraft given the initial and final masses, alongside the exhaust velocity. The equation is given by:\[v_f = v_{ ext{ex}} imes ext{ln} rac{m_i}{m_f}\]where \(v_{ ext{ex}}\) is the exhaust speed, \(m_i\) is the initial mass, and \(m_f\) is the final mass.
02

Calculate the Shuttle's Speed

Apply the values from the problem into the rocket equation:\[v_f = 3000 imes ext{ln} rac{2 imes 10^{6}}{1 imes 10^{6}}\]Calculate the natural logarithm:\[ ext{ln} rac{2 imes 10^{6}}{1 imes 10^{6}} = ext{ln}(2) \]This is approximately 0.693. Therefore:\[v_f = 3000 imes 0.693 = 2079 ext{ m/s}\]The shuttle's speed at the end of this stage is approximately 2079 m/s.
03

Determine the Thrust

Thrust can be calculated using the equation:\[ ext{Thrust} = rac{dm}{dt} imes v_{ ext{ex}}\]The rate of mass expulsion \(\frac{dm}{dt}\) can be found by dividing the total change in mass by the time of expulsion:\[ rac{dm}{dt} = rac{(2 imes 10^{6} - 1 imes 10^{6})}{120 ext{ seconds}} = rac{1 imes 10^{6}}{120}\]Thus the thrust is:\[ ext{Thrust} = rac{1 imes 10^{6}}{120} imes 3000 = 25,000,000 ext{ N}\]The thrust during this period is 25,000,000 N.
04

Compare Thrust with Initial Weight

Calculate the initial weight of the shuttle by multiplying the initial mass by Earth's gravitational acceleration (\(g = 9.81 ext{ m/s}^2\)):\[ ext{Weight}_{ ext{initial}} = 2 imes 10^{6} imes 9.81 = 19,620,000 ext{ N}\]Now compare this with the thrust:\[ rac{ ext{Thrust}}{ ext{Weight}_{ ext{initial}}} = rac{25,000,000}{19,620,000} \]This gives approximately 1.27, indicating the thrust is about 1.27 times the initial weight of the shuttle on Earth.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thrust calculation
Calculating the thrust of a rocket engine is important to understand its capability to lift off from the Earth or maneuver in space. Thrust is generated by expelling mass at high speeds in the opposite direction of desired movement, following Newton's third law of motion: for every action, there is an equal and opposite reaction.

The formula for thrust (\[\text{Thrust} = \frac{dm}{dt} \times v_{\text{ex}}\]), involves two key components:
  • Mass expulsion rate (\(\frac{dm}{dt}\)): This is the rate at which the rocket expels its propellant mass during flight.
  • Exhaust velocity (\(v_{\text{ex}}\)): The speed at which propellant leaves the rocket engine.
For example, if a rocket expels gas at a rate of 1 million kilograms over 120 seconds with an exhaust velocity of 3000 m/s, the calculated thrust would be 25,000,000 N. This high thrust allows the rocket to overcome Earth's gravitational pull and move upward.
Space shuttle launch
The launch of a space shuttle is a carefully orchestrated event that involves the synchronization of several complex systems. A key objective during launch is to overcome Earth's gravity and achieve orbital velocity.

During the initial phase of launch, a shuttle's powerful engines are ignited to produce sufficient thrust to balance and eventually surpass the gravitational pull, allowing ascent into space. The structural integrity, aerodynamic design, and precision timing all play a critical role to ensure safety and efficiency.
  • Shuttles start at massive initial masses, often exceeding millions of kilograms due to the fuel and structures required for the journey.
  • As launch proceeds, fuel is consumed, reducing the total mass and thus increasing acceleration.
  • The contrast between the initial total weight and provided thrust is a crucial factor that determines a successful launch.
The calculation of thrust, as seen with an initial mass of 2 million kg producing thrust well above its weight, demonstrates the capability to initiate liftoff safely.
Exhaust velocity
Exhaust velocity refers to the speed at which gases are expelled from a rocket engine. It is a fundamental aspect that dictates the efficiency and performance of a rocket.

Higher exhaust velocities generally translate to more efficient engines, as they can achieve greater thrust for a given rate of mass expulsion. In the context of the rocket equation, exhaust velocity (\(v_{\text{ex}}\)) is directly proportional to the overall change in velocity (\(v_f\)) of the spacecraft.
  • An average exhaust velocity of 3000 m/s, like in the example, allows the rocket to gain significant speed within a short timeframe.
  • By optimizing the exhaust velocity, engineers can enhance rocket propulsion while managing fuel consumption effectively.
Exhaust velocity, therefore, plays a vital role in determining how quickly a shuttle can ascend and how much speed it can ultimately attain, impacting the mission's success directly.
Mass expulsion rate
The mass expulsion rate is the rate at which a rocket expels its fuel or propellant. This factor is essential for calculating thrust, and it influences the rocket's velocity as described in the rocket equation.

In our context, a mass expulsion rate is calculated by dividing the change in mass (\(\Delta m\)) over a specific time period (\(\Delta t\)):\[\frac{dm}{dt} = \frac{\Delta m}{\Delta t}\]
  • For example, if the shuttle reduces its mass by 1 million kg in 120 seconds, the mass expulsion rate is approximately 8333 kg/s.
  • This measure helps determine how efficiently a rocket uses its propellant, which is crucial for achieving desired speeds and thrust effectively.
Understanding the mass expulsion rate is critical for optimizing flight performance and ensuring that the shuttle can reach its intended destination while maintaining fuel efficiency.

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Most popular questions from this chapter

A uniform thin sheet of metal is cut in the shape of a semicircle of radius \(R\) and lies in the \(x y\) plane with its center at the origin and diameter lying along the \(x\) axis. Find the position of the CM using polar coordinates. [In this case the sum (3.9) that defines the CM position becomes a two- dimensional integral of the form \(\int \mathbf{r} \sigma d A\) where \(\sigma\) denotes the surface mass density (mass/area) of the sheet and \(d A \text { is the element of area } d A=r d r d \phi .]\)

A rocket (initial mass \(m_{\mathrm{o}}\) ) needs to use its engines to hover stationary, just above the ground. (a) If it can afford to burn no more than a mass \(\lambda m_{\mathrm{o}}\) of its fuel, for how long can it hover? [Hint: Write down the condition that the thrust just balance the force of gravity. You can integrate the resulting equation by separating the variables \(t\) and \(m .\) Take \(v_{\text {ex }}\) to be constant.] (b) If \(v_{\mathrm{ex}} \approx 3000 \mathrm{m} / \mathrm{s}\) and \(\lambda \approx 10 \%,\) for how long could the rocket hover just above the earth's surface?

A juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and throws it up so that, at the moment of release, it is horizontal, its \(\mathrm{CM}\) is traveling vertically up at speed \(v_{\mathrm{o}}\) and it is rotating with angular velocity \(\omega_{\mathrm{o}} .\) To catch it, he wants to arrange that when it returns to his hand it will have made an integer number of complete rotations. What should \(v_{\mathrm{o}}\) be, if the rod is to have made exactly \(n\) rotations when it returns to his hand?

In the early stages of the Saturn V rocket's launch, mass was ejected at about \(15,000 \mathrm{kg} / \mathrm{s}\), with a speed \(v_{\mathrm{ex}} \approx 2500 \mathrm{m} / \mathrm{s}\) relative to the rocket. What was the thrust on the rocket? Convert this to tons (1 ton \(\approx 9000\) newtons) and compare with the rocket's initial weight (about 3000 tons).

A uniform spherical asteroid of radius \(R_{\mathrm{o}}\) is spinning with angular velocity \(\omega_{\mathrm{o}}\). As the aeons go by, it picks up more matter until its radius is \(R\). Assuming that its density remains the same and that the additional matter was originally at rest relative to the asteroid (anyway on average), find the asteroid's new angular velocity. (You know from elementary physics that the moment of inertia is \(\frac{2}{5} M R^{2}\).) What is the final angular velocity if the radius doubles?

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